ibmvtpm byteswapping inconsistency

[Michal Suchanek]

On 27 January 2017 at 02:50, Benjamin Herrenschmidt
<benh at kernel.crashing.org> wrote:
> On Thu, 2017-01-26 at 17:42 -0800, Tyrel Datwyler wrote:
>> On 01/26/2017 12:22 PM, Michal Suchánek wrote:
>> > Hello,
>> >
>> > building ibmvtpm I noticed gcc warning complaining that second word
>> > of
>> > struct ibmvtpm_crq in tpm_ibmvtpm_suspend is uninitialized.
>> >
>> > The structure is defined as
>> >
>> > struct ibmvtpm_crq {
>> >         u8 valid;
>> >         u8 msg;
>> >         __be16 len;
>> >         __be32 data;
>> >         __be64 reserved;
>> > } __attribute__((packed, aligned(8)));
>> >
>> > initialized as
>> >
>> >         struct ibmvtpm_crq crq;
>> >         u64 *buf = (u64 *) &crq;
>> > ...
>> >         crq.valid = (u8)IBMVTPM_VALID_CMD;
>> >         crq.msg = (u8)VTPM_PREPARE_TO_SUSPEND;
>> >
>> > and submitted with
>> >
>> >         rc = ibmvtpm_send_crq(ibmvtpm->vdev, cpu_to_be64(buf[0]),
>> >                               cpu_to_be64(buf[1]));
>>
>> These should be be64_to_cpu() here. The underlying hcall made by
>> ibmvtpm_send_crq() requires parameters to be in cpu endian unlike the
>> RTAS interface which requires data in BE.
>
> Hrm... an hcall takes register arguments. Register arguments don't have
> an endianness.
>
> The problem is that we are packing an in-memory structure into 2
> registers and it's expected that this structure is laid out in the
> registers as if it had been loaded by a BE CPU.
>
> So we have two things at play here:
>
>   - The >8-bit fields should be laid out BE in the memory image
>   - That whole 128-bit structure should be loaded into 2 64-bit
> registers MSB first.
>
> So the "double" swap is somewhat needed. The uglyness comes from the
> passing-by-register of the h-call but it should work.
>
> That said, be64_to_cpup(buf) and be64_to_cpup(buf+1) might give you
> better result (though recent gcc's might not make a difference).

If this should work then the below case that swaps the fields separately is
broken.

Anyway, structures have no endianess so when they start with a byte they
start with that byte no matter the host endian.
crq.valid is the first byte always. And then each field is to be swapped
separately.

On the other hand, bitfields are part of an integer and the field should be
swapped as part of the integer.

That is,
#define CRQ_VALID ((buf[0] >> 56) & 0xff)
CRQ_VALID is part of an integer in buf and would be laid out differently
on start or end depending on the host being BE or LE.

And the question is what the PAPR actually defines because both ways are
used in the code. You can describe an in-memory layout either way.

>> >
>> > which means that the second word indeed contains purely garbage.
>> >
>> > This is repeated a few times in the driver so I added memset to
>> > quiet
>> > gcc and make behavior deterministic in case the unused fields get
>> > some
>> > meaning in the future.
>> >
>> > However, in tpm_ibmvtpm_send the structure is initialized as
>> >
>> >     struct ibmvtpm_crq crq;
>> >         __be64 *word = (__be64 *)&crq;
>> > ...
>> >         crq.valid = (u8)IBMVTPM_VALID_CMD;
>> >         crq.msg = (u8)VTPM_TPM_COMMAND;
>> >         crq.len = cpu_to_be16(count);
>> >         crq.data = cpu_to_be32(ibmvtpm->rtce_dma_handle);
>> >
>> > and submitted with
>> >
>> >     rc = ibmvtpm_send_crq(ibmvtpm->vdev, be64_to_cpu(word[0]),
>> >                               be64_to_cpu(word[1]));
>> > meaning it is swapped twice.
>> >
>> >
>> > Where is the interface defined? Are the command arguments passed as
>> > BE
>> > subfields (the second case was correct before adding the extra
>> > whole
>> > word swap) or BE words (the first case doing whole word swap is
>> > correct)?
>>
>> The interface is defined in PAPR. The crq format is defined in BE
>> terms.

Which exact PAPR? Where can I get it?
The PAPR document I found does not say anything about vtpm.

Thanks

Michal